The document NCERT Solutions (Part- 4)- Algebraic Expressions and Identities Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.

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**Exercise 9.5**

**Question 1. **Use a suitable identity to get each of the following products.

**Solution: (i)** Using the identity,

(x + a)(x + b) = x^{2} + (a + b)x + ab

we have

(x + 3)(x + 3) = x^{2} + (3 + 3)x + (3 * 3)

= x^{2} + 6x + 9

**(ii)** Using the identity,

(x + a)(x + b) = x^{2} + (a + b)x + ab,

we have

(2y + 5)(2y + 5) = (2y)^{2} + (5 + 5)2y + (5 * 5)

= 4y^{2} + (10)2y + 25

= 4y^{2} + 20y + 25

**(iii)** Using the identity, (a â€“ b)^{2} = a^{2} â€“ 2ab + b^{2}

(2a â€“ 7)(2a â€“ 7) = (2a â€“ 7)^{2}

= (2a)^{2} â€“ 2(2a)(7) + (7)^{2}

= 4a2 â€“ 28a + 49

**(iv)** Using the Identity, (a â€“ b)^{2} = a^{2} â€“ 2ab + b^{2}, we have

**(v)** Using the identity, (a + b)(a â€“ b) = a^{2} â€“ b^{2},

we have

(1.1m â€“ 0.4)(1.1m â€“ 0.4) = (1.1m)^{2} â€“ (0.4)^{2}

= 1.21m^{2} â€“ 0.16

**(vi)** âˆµ (a^{2} + b^{2})(â€“a^{2} + b^{2}) = (b^{2} + a^{2})(b^{2} â€“ a^{2})

âˆ´ Using the identity, (a + b)(a â€“ b) = a^{2} â€“ b^{2}, we have

(b^{2} + a^{2})(b^{2} â€“ a^{2})= (b^{2})^{2} â€“ (a^{2})^{2} = b^{4} â€“ a^{4}

**(vii)** âˆµ (6x â€“ 7)(6x + 7) = (6x + 7)(6x â€“ 7)

âˆ´ Using the identity, (a + b)(a â€“ b) = a^{2} â€“ b^{2},

we have

(6x + 7)(6x â€“ 7) = (6x)^{2} â€“ (7)^{2}

= 36x^{2} â€“ 49

**(viii)** Using the identity, (a + b)^{2} = a^{2} + 2ab + c^{2}, we have

(â€“a + c)(â€“a + c) = (â€“a + c)^{2}

= (â€“a)2 + 2(â€“a)(c) + (c)^{2}

= a^{2} + 2(â€“ac) + c^{2}

= a^{2} â€“ 2ac + c^{2}

**(ix)** Using the identity, (a + b)^{2} = a^{2} + b^{2} + 2ab, we have

**(x) **Using the identity, (a â€“ b)^{2} = a^{2} â€“ 2ab + b^{2} we have

(7a â€“ 9b)(7a â€“ 9b) = (7a â€“ 9b)^{2}

= (7a)^{2} â€“ 2(7a)(9b) + (9b)^{2}

= 49a^{2} â€“ 126ab + 81b^{2}

**Question 2.** Use the identity (x + a)(x + b) = x^{2} + (a + b)x + ab to find the following products.

**(i)** (x + 3)(x + 7) **(ii) **(4x + 5)(4x + 1) **(iii)** (4x â€“ 5)(4x â€“ 1)**(iv)** (4x + 5)(4x â€“ 1) **(v)** (2x + 5y)(2x + 3y)**(vi)** (2a^{2} + 9)(2a^{2} + 5) **(vii) **(xyz â€“ 4)(xyz â€“ 2)

**Solution: **

**Question 3.** Find the following squares by using the identities.

**(i)** (b â€“ 7)^{2 } **(ii) **(xy + 3z)^{2} **(iii) **(6x^{2} â€“ 5y)^{2 }

**(v)** (0.4p â€“ 0.5q)^{2} **(vi) **(2xy + 5y)^{2}

**Solution: **

**(i)** (b â€“ 7)^{2}

Using (a â€“ b)^{2 }= a^{2} â€“ 2ab â€“ b^{2}, we have

(b â€“ 7)^{2} = (b)^{2} â€“ 2(b)(7) + (7)^{2}

= b^{2} â€“ 14b + 49

**Question 4.** Simplify:

**(i)** (a^{2} â€“ b^{2})^{2 }** (ii) **(2x + 5)^{2 }â€“ (2x â€“ 5)^{2}

**(iii)** (7m â€“ 8n)2 + (7m + 8n)^{2} **(iv)** (4m + 5n)^{2} + (5m + 4n)^{2}

**(v)** (2.5p â€“ 1.5q)^{2} â€“ (1.5p â€“ 2.5q)^{2} **(vi)** (ab + bc)^{2} â€“ 2ab^{2}c

**(vii) **(m^{2} â€“ n^{2}m)^{2} + 2m^{3}n^{2}

**Solution:**

(i) (a^{2} â€“ b^{2})^{2} = (a^{2})^{2} â€“ 2a^{2} * b^{2} + (b^{2})^{2} [using (a â€“ b)^{2} = a^{2} â€“ 2ab + b^{2}]

= a^{4} â€“ 2a^{2}b^{2} + b^{4}

**Question 5.** Show that

**Solution:**

**(i) **LHS = (3x + 7)^{2} â€“ 84x

= (3x)^{2} + 2(3x)(7) + (7)^{2} â€“ 84x

= 9x^{2} + 42x + 49 â€“ 84x

= 9x^{2} + (42 â€“ 84)x + 49

= 9x^{2} â€“ 42x + 49

RHS = (3x â€“ 7)^{2}

= (3x)^{2} â€“ 2(3x)(7) + (7)^{2}

= 9x^{2} â€“ 42x + 49 = LHS

Since, LHS = RHS

âˆ´ (3x + 7)^{2} â€“ 84x = (3x â€“ 7)^{2}

**(ii) **LHS = (9p â€“ 5q)^{2 }+ 180pq

= (9p)^{2} â€“ 2(9p)(5q) + (5q)^{2} + 180pq

= 81p^{2} â€“ 90pq + 25q^{2} + 180pq

= 81p^{2 }+ (â€“90 + 180)pq + 25q^{2}

= 81p^{2 }+ 90pq + 25q^{2}

RHS = (9p + 5q)^{2}

= (9p)^{2} + 2(9p)(5q) + (5q)^{2}

= 81p^{2} + 90pq + 25q^{2}

Since, LHS = RHS

âˆ´ (9p â€“ 5q)^{2} + 180pq = (9p + 5q)^{2}

**Question 6.** Using identities, evaluate:

(i) 71^{2} (ii) 99^{2} (iii) 102^{2} (iv) 998^{2} (v) 5.2^{2} (vi) 297 * 303 (vii) 78 * 82 (viii) 8.9^{2} (ix) 1.05 * 9.5

**Solution: **

**(i) **âˆµ (71) = (70 + 1)

âˆ´ (71)2 = (70 + 1)^{2}

= (70)^{2} + 2(70)(1) + (1)^{2} [Using (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 4900 + 140 + 1 = 5041

**(ii) **We have 100 â€“ 1 = 99

âˆ´ (99)^{2} = (100 â€“ 1)^{2}

= (100)^{2} â€“ 2(100)(1) + (1)^{2} [Using (a â€“ b)^{2} = a^{2} â€“ 2ab + b^{2}]

= 10000 â€“ 200 + 1 = 9801

**(iii) **âˆµ 102 = 100 + 2

âˆ´ (102)^{2} = (100 + 2)^{2}

= (100)^{2} + 2(100)(2) + (2)^{2} [Using (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 10000 + 400 + 4 = 10404

**(iv)** âˆµ 998 = 1000 â€“ 2

âˆ´ (998)^{2} = (1000 â€“ 2)^{2}

= (1000)^{2} â€“ 2(1000)(2) + (2)^{2} [Using (a â€“ b)^{2} = a^{2} â€“ 2ab + b^{2}]

= 1000000 â€“ 4000 + 4 = 996004

**(v)** âˆµ 5.2 = 5 + 0.2

âˆ´ (5.2)^{2} = (5 + 0.2)^{2}

= (5)^{2} + 2(5)(0.2) + (0.2)^{2} [Using (a + b)^{2} = a^{2} + 2ab + b^{2}]

= 25 + 2 + 0.04 = 27.04

**(vi)** âˆµ 297 = 300 â€“ 3 and 303 = 300 + 3

âˆ´ 297 * 303 = (300 â€“ 3)(300 + 3)

= (300)^{2} â€“ (3)^{2} [Using (a + b)(a â€“ b) = a^{2} â€“ b2]

= 90000 â€“ 9 = 89991

**(vii) **âˆµ 78 = 80 â€“ 2 and 82 = 80 + 2

âˆ´ 78 * 82 = (80 â€“ 2)(80 + 2)

= 80^{2} â€“ (2)^{2} [Using (a + b)(a â€“ b) = a^{2} â€“ b^{2}]

= 6400 â€“ 4 = 6396

**(viii)** âˆµ 8.9 = (9 â€“ 0.1)

âˆ´ (8.9)^{2} = (9 â€“ 0.1)

= (9)^{2} â€“ 2(9)(0.1) + (0.1)^{2} [Using (a â€“ b)^{2} = a^{2} â€“ 2ab + b^{2}]

= 81 â€“ 1.8 + 0.01

= 81.01 â€“ 1.80 = 79.21

**(ix) **âˆµ 1.05 = 1 + 0.05

âˆ´ (1.05) * 9.5 = (1 + 0. 05) * 9.5

= (1 * 9.5) + (9.5 * 0.05)

= 9.500 + 0.475 = 9.975

**Question 7.** Using a^{2} â€“ b^{2} = (a + b)(a â€“ b), find:

(i) 51^{2} â€“ 49^{2} (ii) (1.02)^{2 }â€“ (0.98)^{2} (iii) 153^{2} â€“ 147^{2} (iv) 12.1^{2} â€“ 7.9^{2}

**Solution: **

**(i) **51^{2} â€“ 49^{2} = (51 + 49)(51 â€“ 49)

= (100)* (2) = 200

**(ii) **(1.02)^{2} â€“ (0.98)^{2} = (1.02 + 0.98)(1.02 â€“ 0.98)

= (2.0) * (0.04) = 0.08

**(iii)** 153^{2} â€“ 147^{2} = (153 + 147)(153 â€“ 147)

= (300) * (6) = 1800

**(iv) **(12.1)^{2} â€“ (7.9)^{2} = (12.1 + 7.9)(12.1 â€“ 7.9)

= 20 * 4.2 = 84

**Question 8.** Using (x + a)(x + b) = x^{2} + (a + b)x + ab,

find: (i) 103 * 104 (ii) 5.1 * 5.2

(iii) 103 * 98 (iv) 9.7 * 9.8

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